Next picture please. Now I have a very strange cutting here. Where we take the vector equilibrium, and we take the one-eight octahedron that we put onto each of the triangular faces to make it into a cube. Then, you remember that the vector equilibrium has, as a sphere, it had 25 great circles, remember that. And remember those 25 great circles, there were the 3 great circles, a 4 great circles that's 7. There was a 6 great circles makes 13. And there were 12 great circles, made a total of 25. If we had each one of those great circle planes were extended into the cube, and if I had those planes, then, come out I've got this corner over here, and they get outside the vector equilibrium and they cut up this little corner into pieces. Go back to that picture, may I see that picture again?

So what you see there on the lower right are the one-eighth octahedron corner. And you can see the lines on the surface of it, and it has been cut up internally by all the twenty five great circles. Now, back in 1947, I did my trigonometry carefully, and I found what the volumes of these were and they all counted out rational number. These little fractions are coming out rational numbers! Boy, you'd better work harder still. And, I'd just like to have that one mentioned.

Next picture. Now here is the, looking at the vector equilibrium, cut through the center, and we'll see a central vector equilibrium, and one, two, three there are four enclosures. I found that whereas I gave you for each layer in terms of spheres there were 12, 42, 92 and so forth, I found that the volumes were growing. The volume of the vector equilibrium is 20. And the sum total volume of the vector equilibrium is always frequency to the third power times twenty. For instance if the frequency were twenty then the volume would be three to the fourth power. This is a very we do not start with a zero, we start with an entity twenty is unity. Unity can never be less in a vector equilibrium unity is twenty. In the octahedron, unity is four. It never is less. That's where the number begins. So that when we, I found that the rate at which layers grew of the vector equilibrium, and I gave you the series of layers. If I took where the outer set of balls were occurring between that ball and the next layer in it, there would be a plane that could be struck; and that for any two layers, they combined to give me a very interesting number. Where the volumes were growing at frequency to the second power times six plus two. In other words there was something to do with this spheric space this number 6 being very unique to the rhombic dodecahedron or the domain of the sphere. So that those spaces that you're looking at now, pairing any two, will always be frequency to the second power times six plus two.

Next picture. What we are looking at here in this lower right hand corner is quite difficult to see. You are looking at the icosahedron, and it is you have the perpendicular bisectors of it, in other words we're looking at the 15 great circles which give you 120 spaces 120 right triangles on the surface of the sphere which I said the Babylonian's identified as very important. If I could have the picture back again. I found then that insomuch as the octahedron has eight faces, and there were 120 triangular faces showing there, I said, "Can you divide 120 by 8?" and I found you could. It's 15. And I said "It could be that if I took the icosahedron's face, and added around each of the three edges I would add, because I know that there are six inside, there are six triangles inside the icosahedron's face, so 6 from 15 leaves me 9. There are 3 edges. So therefore, if I could find a way of mounting 3 triangles on each edge of the icosahedron's triangle, in a symmetrical way, where it would make a corner of 90 degrees, like the octahedron, then I might have found an octahedron inside the icosahedron and sure enough we do have it. On the lower right hand side you will then see the octahedron in the icosahedron. They skew like this. Tomorrow I'll bring a model here where you'll see the icosahedron inside the octahedron and you'll find that literally it is rotatable. Will you bring that, Meddy, from the office tomorrow.

Next picture. Now, remember, I am looking at, again I'm not seeing enough. I'm going to go to the board to show you your squares, and triangles and your counts were all perfectly clear; but we have the two triangles to the square but I want to show you something about triangles and squares, which then also applies to rectilinears. (Bucky goes to the board now). That model is here. Oh beautiful! We won't go any further without it. You recognize your yellow octahedron, now, and your red icosahedron. And notice, it is skew in there. The points like that. And I'm going to do this. It will rotate right in there. Fantastic. The beauty of these two coming together. So I want you to get into the rationality of the interrelationship of this icosahedron, which is always bothering you because it's volume is 18.51 all the others are rational. And you have to realize it is a very special behavior, that icosahedron, because it can only have one layer of the balls in closest packing due to the contraction there is no room for the second layer. The vector equilibrium position there is room for the layers, but when it is contracted, there is only room for the outer layer the outer layer can be triangulated, but you can have no more layers, because it is inherently a single-layer affair. So, again, it makes me feel that it is very much the electron behavior, and I have to note these kind of reciprocities.

Now, I was going to come over to talking about. If I make a any quadrangular form, but not equal edges if I bisect those edges and interconnect, I do not get four similar forms. That's not so. But if I take any triangle and bisect its edges and interconnect, I get four exactly similar triangles. That is, the triangle is inviolable in the matter of its it can look like anything to you in this Universe, but the accountings I have been giving to you always come out exactly the same whether it looks like a regular tetrahedron or not.

And the same thing if I made that a rectilinear thing, it would be a whole lot of trouble as you know. So that but in the tetrahedron can look like anything you want, and it still comes out this way. Now, there were two forms that showed in the last pictures where I saw that they were not clear enough to really bring to your attention. But I now am going to get into an accounting, just doing this for you myself.

I'm going to make a I've got a tetrahedron made I've got rubber bands coming from me, and I'm going to hold the base of my tetrahedron between my two feet. This is a basic one. And you can take a hold of the vertex over there, and I am going to make a line parallel to my feet over her. And I'm going to hold this tetrahedron's top here, as long as you just move it in this line always parallel to me, the area of the triangle it forms with the base here, will be the same base and the same altitude, because the lines are parallel to one another. Agreed? So then I find that I can move that base line and I also can move my top here in a plane parallel to the floor. So I find that out of the tetrahedron's four corners I hold two of them, I hold one edge fixed only. And the other five edges are continually changing. But it always remains the same size tetrahedron. So I can move this vertex way, way over there. Or I can move that one on the floor over there. Always the same altitude, the same base. In other words, you could get to an extraordinarily asymmetrical tetrahedron here, but it always holds true. And I am quite confident that this has a whole lot to do with the fundamentals of tuning. Where I said, I could sub-divide those triangles any way I want. It doesn't make any difference at all. It all comes out, always, the same value. I think it's a whole lot to do with electromagnetic tuning, where you could tune in here, and I could tune in any place in the Universe and we simply come in. I haven't made that particularly clear that point, but as time goes on, that will clarify.

Next picture please. On the first day I tried to have a stick standing up here and it didn't work. Then I had a picture over here, but it didn't really do much better. I've got two sticks standing in the lower left hand side, and they can swerve anywhere, but over on the right they fall towards each other. And so they engaged these other tops, and now they can act as a hinge.

Next picture. And here we have the three of them falling towards each other and making the tetrahedron. But it's legs can go out, so we must have a set of tensions down there.

Next picture. That's part of what I've just been talking about.

Next picture please. About the distortions being in there. Here you have a drawing I was making of tetrahedra. These very simple things where they get very narrow, and get to be like the tree, so I gave you the other day a tree as we know, the cambium layer. Each year the cambium layer encloses the next one, and we find they literally do grow, the top is now, they are tetrahedronal. They might look more like a cone to start off with, but, no, you'll find the roots that as they do the stretches tend to go out tetrahedronally. So that I said each layer of the cambium layer the trees continue, one tetrahedron enclosing another, and the branches themselves go out as tetrahedra, and you find the wing root is really exactly that, the bottom of the wing root in here it's deep, and the flap goes that way that's the shape of the wing root. So that each branch is a tetrahedron, and it gets its cambium layers, and each twig is the same, and they are continually this way with tetrahedron embracing tetrahedron, and so the bud and the whole thing comes out there.